Lecture Notes on Operators and Eigenvalue Problems in Quantum Mechanics

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February 5, 2025

Introduction

This lecture addresses essential concepts in quantum mechanics, focusing on operators and eigenvalue problems. We begin by rectifying a mistake from the previous lecture, followed by an introduction to the change of basis and unitary operators. We emphasize the importance of unitary operators in quantum mechanics, particularly in describing time evolution. Subsequently, we define and discuss various types of operators, including unitary, normal, and Hermitian operators, highlighting their properties and significance. A significant portion of this lecture is dedicated to the eigenvalue problem, covering its formulation, the secular equation, and the concept of eigenvectors. We explore the properties of eigenvectors, especially for normal operators, and introduce the spectral representation of operators. Finally, we discuss special operators such as Hermitian and unitary operators, their unique eigenvalue characteristics, the trace of an operator, functions of operators, and tensor products for composite systems.

Correction from Previous Lecture

We begin by addressing an error from the previous lecture concerning index manipulation in a demonstration. The correction is presented in these slides for clarity and ease of understanding.

Change of Basis and Unitary Operators

In vector spaces, a change of basis can be described by operators, denoted here as $R$, that satisfy the following conditions: \[R R^\dagger = I \quad \text{and} \quad R^\dagger R = I, \label{eq:unitary_condition}\] where $I$ is the identity operator. This is analogous to rotations in linear algebra, where the inverse of a rotation $R$ is $R^{-1}$, and in this context, the adjoint operator $R^\dagger$ serves as the inverse, i.e., $R^\dagger = R^{-1}$. These operators, which include rotations and reflections, are characterized by their property of preserving the length or norm of vectors.

Definition 1. A transformation $U$ is termed a unitary transformation or unitary operator if it satisfies the condition: \[U U^\dagger = U^\dagger U = I. \label{eq:unitary_operator_definition}\] In quantum computing, such operators are often referred to simply as unitaries.

Unitary operators are crucial because they preserve the norm of vectors, a property that is essential in quantum mechanics for maintaining the conservation of probabilities under transformations.

Importance of Unitary Operators in Quantum Mechanics

Unitary operators are of fundamental importance in quantum mechanics as they describe the time evolution of quantum systems. In the context of quantum circuits, the evolution of a quantum state as it passes through the circuit is represented by a sequence of unitary operators. Each quantum gate in a circuit corresponds to a unitary transformation. Consequently, any quantum circuit can be mathematically represented by a composition of unitary matrices.

Furthermore, unitary operators can be interpreted geometrically as rotations in Hilbert space, effectively changing from one orthonormal basis to another. This perspective is valuable for visualizing and understanding the transformations applied to quantum states.

Operators in Quantum Mechanics

Unitary Operators

Definition and Properties

Definition 2. An operator $U$ is defined as unitary if its adjoint $U^\dagger$ is also its inverse, satisfying the condition: \[U U^\dagger = U^\dagger U = I, \label{eq:unitary_formal_definition}\] where $I$ represents the identity operator.

Unitary operators are crucial in quantum mechanics as they guarantee that transformations are both reversible and probability-preserving. This reversibility ensures that quantum evolution can be traced back, and the probability preservation is essential for maintaining the probabilistic interpretation of quantum states.

Unitary Operators as Time Evolution Operators

Unitary operators play a central role in quantum mechanics by describing the time evolution of isolated quantum systems. The transition of a quantum state from time $t_1$ to time $t_2$ is mathematically described by the application of a unitary operator. This concept is fundamental to both quantum dynamics and quantum computation, where sequences of unitary operators represent quantum algorithms.

Normal Operators

Definition

Definition 3. An operator $A$ is termed a normal operator if it commutes with its adjoint $A^\dagger$. The condition for normality is given by: \[A A^\dagger = A^\dagger A. \label{eq:normal_operator_definition}\]

Normal operators are significant because of their desirable spectral properties. Specifically, they are diagonalizable, and their eigenvectors are guaranteed to form an orthonormal basis for the Hilbert space on which they act. This property simplifies many theoretical and computational tasks in quantum mechanics.

Hermitian Operators

Definition

Definition 4. An operator $H$ is defined as a Hermitian operator if it is equal to its adjoint, i.e., \[H = H^\dagger. \label{eq:hermitian_operator_definition}\]

Hermitian operators constitute an important subclass of normal operators. If an operator $H$ is Hermitian ($H = H^\dagger$), it directly follows that $H H^\dagger = H H = H^\dagger H$, thus satisfying the commutative property required for normal operators as defined in [eq:normal_operator_definition].

Physical Observables and Hermitian Operators

In quantum mechanics, a fundamental postulate is that every physical observable quantity—such as energy, momentum, position, and spin—is represented by a Hermitian operator. This correspondence is a cornerstone of quantum theory, linking mathematical operators to measurable physical properties.

Remark. Remark 1. It is important to note that while every physical observable is represented by a Hermitian operator, the converse is not necessarily true. Not every Hermitian operator corresponds to a directly observable physical quantity. However, the set of all physical observables is always a subset of the set of Hermitian operators.

Eigenvalue Problem

Introduction to the Eigenvalue Problem

The eigenvalue problem is a cornerstone of linear algebra and quantum mechanics. It aims to identify specific vectors, termed eigenvectors, which, when acted upon by a linear operator, are merely scaled by a factor, known as the eigenvalue.

Mathematically, for a given operator $A$ and a non-zero vector $\left|{v}\right\rangle$, the eigenvalue problem is formulated as finding scalars $\lambda \in \mathbb{C}$ and vectors $\left|{v}\right\rangle \neq 0$ that satisfy: \[A \left|{v}\right\rangle = \lambda \left|{v}\right\rangle. \label{eq:eigenvalue_equation}\] In this equation, $\lambda$ is the eigenvalue, and $\left|{v}\right\rangle$ is the corresponding eigenvector of the operator $A$.

Secular Equation and the Characteristic Polynomial

To solve the eigenvalue problem, we rearrange [eq:eigenvalue_equation] into the form: \[(A - \lambda I) \left|{v}\right\rangle = 0, \label{eq:secular_equation_operator}\] where $I$ is the identity operator. For [eq:secular_equation_operator] to have non-trivial solutions (i.e., $\left|{v}\right\rangle \neq 0$), the operator $(A - \lambda I)$ must be singular, meaning it must not have an inverse. This condition is satisfied if and only if the determinant of $(A - \lambda I)$ is zero. This condition leads to the secular equation or characteristic equation: \[\det(A - \lambda I) = 0. \label{eq:secular_equation_determinant}\] Solving [eq:secular_equation_determinant] for $\lambda$ yields the eigenvalues of the operator $A$. The expression $\det(A - \lambda I)$ is a polynomial in $\lambda$, known as the characteristic polynomial.

Rank, Linear Dependence, and the Determinant

The determinant of a square matrix is intrinsically linked to the concept of linear independence of its rows (or columns) and the rank of the matrix.

Definition 5. The rank of a matrix is defined as the maximum number of linearly independent rows (or columns) in the matrix. For an $n \times n$ square matrix, the rank can range from 0 to $n$. A matrix is said to have full rank if its rank is equal to its dimension $n$.

A crucial property is that a square matrix has a determinant of zero if and only if its rows (or columns) are linearly dependent, which is equivalent to the matrix not having full rank. In other words, if the rank of an $n \times n$ matrix is less than $n$, its determinant is zero. Conversely, a non-zero determinant implies that the matrix has full rank and its rows and columns are linearly independent.

Definition 6. The determinant of an $n \times n$ matrix $M$, denoted as $\det(M)$, is formally defined using permutations as: \[\det(M) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^{n} M_{i, \sigma(i)}, \label{eq:determinant_definition}\] where $S_n$ is the set of all permutations of $\{1, 2, \ldots, n\}$, and $\text{sgn}(\sigma)$ is the sign of the permutation $\sigma$.

While [eq:determinant_definition] provides a formal definition, for matrices of larger dimensions, calculating the determinant directly using this formula becomes computationally intensive due to the factorial growth of the number of permutations ($n!$). In practice, more efficient algorithms, such as Gaussian elimination or LU decomposition, are employed for determinant computation. However, for theoretical understanding, the definition in [eq:determinant_definition] and its connection to linear dependence are fundamental.

Existence of Eigenvalues and the Fundamental Theorem of Algebra

The secular equation, $\det(A - \lambda I) = 0$, derived from the eigenvalue problem, is a polynomial equation in $\lambda$. For an $n \times n$ matrix $A$, this equation is a polynomial of degree $n$. The Fundamental Theorem of Algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. Furthermore, a polynomial equation of degree $n$ has exactly $n$ complex roots, counted with multiplicity.

Therefore, for any $n \times n$ matrix (representing an operator in an $n$-dimensional vector space), there exist precisely $n$ eigenvalues. These eigenvalues are complex numbers in general, although they may be real in specific cases (e.g., for Hermitian operators). Eigenvalues can also be degenerate, meaning that some eigenvalues may be repeated roots of the characteristic polynomial.

Eigenvectors and Normalization

Once the eigenvalues $\{\lambda_i\}$ are determined by solving the secular equation, the corresponding eigenvectors $\{\left|{v_i}\right\rangle\}$ can be found by solving the homogeneous system of linear equations given by [eq:secular_equation_operator] for each $\lambda_i$: \[(A - \lambda_i I) \left|{v_i}\right\rangle = 0. \label{eq:eigenvector_equation}\] For each eigenvalue $\lambda_i$, [eq:eigenvector_equation] guarantees at least one non-zero solution $\left|{v_i}\right\rangle$, which is the eigenvector associated with $\lambda_i$. It is important to note that eigenvectors are defined up to a non-zero multiplicative constant. To standardize eigenvectors and for convenience in many applications, it is common practice to normalize them such that they have unit length, i.e., $\left\langle{v_i}\middle|{v_i}\right\rangle = 1$.

Example: Eigenvalues and Eigenvectors of the Pauli-X Matrix

To illustrate the eigenvalue problem, consider the Pauli-X matrix, a fundamental operator in quantum computing: \[\sigma_x = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \label{eq:pauli_x_matrix}\] To find the eigenvalues of $X$, we solve the secular equation $\det(X - \lambda I) = 0$: \[\det \begin{pmatrix} -\lambda & 1 \\ 1 & -\lambda \end{pmatrix} = (-\lambda)(-\lambda) - (1)(1) = \lambda^2 - 1 = 0. \label{eq:secular_equation_pauli_x}\] The solutions to [eq:secular_equation_pauli_x] are $\lambda_1 = 1$ and $\lambda_2 = -1$, which are the eigenvalues of the Pauli-X matrix.

To find the eigenvector $\left|{v_1}\right\rangle$ corresponding to $\lambda_1 = 1$, we solve $(X - I) \left|{v_1}\right\rangle = 0$: \[\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \label{eq:eigenvector_equation_lambda1}\] This matrix equation yields the condition $-\alpha + \beta = 0$, or $\beta = \alpha$. Choosing $\alpha = 1$ and normalizing the resulting vector, we obtain the normalized eigenvector: \[\left|{v_1}\right\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} (\left|{0}\right\rangle + \left|{1}\right\rangle). \label{eq:eigenvector_v1_pauli_x}\]

Similarly, for the eigenvalue $\lambda_2 = -1$, we solve $(X + I) \left|{v_2}\right\rangle = 0$: \[\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \label{eq:eigenvector_equation_lambda2}\] From [eq:eigenvector_equation_lambda2], we get $\alpha + \beta = 0$, or $\beta = -\alpha$. Choosing $\alpha = 1$ and normalizing, we find the normalized eigenvector: \[\left|{v_2}\right\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} (\left|{0}\right\rangle - \left|{1}\right\rangle). \label{eq:eigenvector_v2_pauli_x}\] Thus, $\left|{v_1}\right\rangle$ and $\left|{v_2}\right\rangle$ are the eigenvectors of the Pauli-X matrix corresponding to the eigenvalues $1$ and $-1$, respectively.

Properties of Eigenvectors and Spectral Representation

Eigenvectors of Normal Operators

Orthogonality of Eigenvectors for Normal Operators

A crucial property of normal operators is the orthogonality of their eigenvectors, especially when the eigenvalues are distinct.

Theorem 1. Eigenvectors of a normal operator corresponding to distinct eigenvalues are mutually orthogonal.

Description: This theorem states that for a normal operator, if we have eigenvectors associated with different eigenvalues, those eigenvectors will always be orthogonal to each other.

Proof. Proof. Let $A$ be a normal operator, and consider two eigenvectors $\left|{v_i}\right\rangle$ and $\left|{v_j}\right\rangle$ corresponding to distinct eigenvalues $\lambda_i$ and $\lambda_j$ respectively, such that $\lambda_i \neq \lambda_j$. By definition, we have: \[\begin{aligned} A \left|{v_i}\right\rangle &= \lambda_i \left|{v_i}\right\rangle, \label{eq:eigenvector_eq_i} \\ A \left|{v_j}\right\rangle &= \lambda_j \left|{v_j}\right\rangle. \label{eq:eigenvector_eq_j}\end{aligned}\] Consider the inner product $\left\langle{v_i}\middle|{A v_j}\right\rangle$. Using [eq:eigenvector_eq_j], we can write: \[\left\langle{v_i}\middle|{A v_j}\right\rangle = \left\langle{v_i}\middle|{\lambda_j v_j}\right\rangle = \lambda_j \left\langle{v_i}\middle|{v_j}\right\rangle. \label{eq:inner_product_Avj}\] Alternatively, we can use the property of the adjoint operator and the normality of $A$ ($A A^\dagger = A^\dagger A$). We know that if $\left|{v_i}\right\rangle$ is an eigenvector of a normal operator $A$ with eigenvalue $\lambda_i$, then $\left|{v_i}\right\rangle$ is also an eigenvector of $A^\dagger$ with eigenvalue $\lambda_i^*$. Thus, $A^\dagger \left|{v_i}\right\rangle = \lambda_i^* \left|{v_i}\right\rangle$. Now, consider $\left\langle{v_i}\middle|{A v_j}\right\rangle$ again, and use the adjoint property: \[\left\langle{v_i}\middle|{A v_j}\right\rangle = \left\langle{A^\dagger v_i}\middle|{v_j}\right\rangle = \left\langle{\lambda_i^* v_i}\middle|{v_j}\right\rangle = \lambda_i \left\langle{v_i}\middle|{v_j}\right\rangle. \label{eq:inner_product_Adag_vi}\] Equating [eq:inner_product_Avj] and [eq:inner_product_Adag_vi], we get: \[\lambda_j \left\langle{v_i}\middle|{v_j}\right\rangle = \lambda_i \left\langle{v_i}\middle|{v_j}\right\rangle.\] Rearranging this equation yields: \[(\lambda_j - \lambda_i) \left\langle{v_i}\middle|{v_j}\right\rangle = 0.\] Since we assumed distinct eigenvalues, $\lambda_i \neq \lambda_j$, which implies $(\lambda_j - \lambda_i) \neq 0$. Therefore, for the equation to hold, we must have: \[\left\langle{v_i}\middle|{v_j}\right\rangle = 0.\] This demonstrates that eigenvectors corresponding to distinct eigenvalues of a normal operator are orthogonal. ◻

Orthonormal Basis from Eigenvectors

For a normal operator acting on an $n$-dimensional vector space, we can always find a set of $n$ eigenvectors that form a basis for this space. According to 1, eigenvectors corresponding to distinct eigenvalues are orthogonal. In cases where eigenvalues are degenerate (i.e., an eigenvalue has multiplicity greater than one), the eigenvectors corresponding to a degenerate eigenvalue span a subspace known as the eigenspace. Within each eigenspace, we can choose a set of orthogonal vectors, for instance, by applying the Gram-Schmidt orthonormalization process if necessary, to form an orthonormal basis for that subspace. Combining the orthonormal bases from all eigenspaces, we obtain a complete orthonormal basis for the entire $n$-dimensional vector space, consisting of eigenvectors of the normal operator.

Spectral Representation of Normal Operators

The spectral representation, also known as the eigenvalue decomposition, provides a way to express a normal operator in terms of its eigenvalues and eigenvectors.

Definition 7. For a normal operator $A$ with a complete set of orthonormal eigenvectors $\{\left|{v_i}\right\rangle\}$ and corresponding eigenvalues $\{\lambda_i\}$, the spectral representation of $A$ is given by: \[A = \sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right| = \sum_{i} \lambda_i P_i, \label{eq:spectral_representation}\] where $P_i = \left|{v_i}\right\rangle \left\langle{v_i}\right|$ is the projection operator onto the eigenvector $\left|{v_i}\right\rangle$.

This representation decomposes the operator $A$ into a sum of projections onto its eigenvectors, each weighted by its corresponding eigenvalue. The spectral representation is particularly useful for understanding the action of an operator and for defining functions of operators.

Example: Spectral Representation of the Pauli-Z Matrix

Consider the Pauli-Z matrix, $\sigma_z = Z$, given by: \[Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \label{eq:pauli_z_matrix}\] The eigenvalues of $Z$ are readily found to be $\lambda_1 = 1$ and $\lambda_2 = -1$. The eigenvector corresponding to $\lambda_1 = 1$ is $\left|{v_1}\right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \left|{0}\right\rangle$, and the eigenvector for $\lambda_2 = -1$ is $\left|{v_2}\right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \left|{1}\right\rangle$. Both eigenvectors are already orthonormal.

Example 1. Using [eq:spectral_representation], the spectral representation of $Z$ is: \[\begin{aligned} Z &= \lambda_1 \left|{v_1}\right\rangle \left\langle{v_1}\right| + \lambda_2 \left|{v_2}\right\rangle \left\langle{v_2}\right| \\ &= (1) \left|{0}\right\rangle \left\langle{0}\right| + (-1) \left|{1}\right\rangle \left\langle{1}\right| \\ &= \left|{0}\right\rangle \left\langle{0}\right| - \left|{1}\right\rangle \left\langle{1}\right|.\end{aligned}\] In matrix form, this is: \[\begin{aligned} Z &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\end{aligned}\] which is indeed the Pauli-Z matrix. This example illustrates how the spectral representation decomposes the Pauli-Z operator in terms of its eigenvalues and the projectors onto its eigenvectors, $\left|{0}\right\rangle\left\langle{0}\right|$ and $\left|{1}\right\rangle\left\langle{1}\right|$.

Special Operators and their Properties

Hermitian Operators

Real Eigenvalues of Hermitian Operators

A fundamental property of Hermitian operators is that their eigenvalues are always real. This is crucial in quantum mechanics because physical observables are represented by Hermitian operators, and their eigenvalues correspond to the possible measured values, which must be real.

Theorem 2. The eigenvalues of a Hermitian operator are real numbers.

Description: This theorem demonstrates that for any Hermitian operator, the eigenvalues obtained by solving the eigenvalue equation will always be real numbers. This is a crucial property in quantum mechanics as physical observables are represented by Hermitian operators, and their measured values must be real.

Proof. Proof. Let $H$ be a Hermitian operator, and let $\lambda$ be an eigenvalue of $H$ with corresponding eigenvector $\left|{v}\right\rangle \neq 0$. By definition, this means $H \left|{v}\right\rangle = \lambda \left|{v}\right\rangle$. To prove that $\lambda$ is real, we start by considering the expectation value of $H$ in the state $\left|{v}\right\rangle$, which can be written in two ways:

First, using the eigenvalue equation: \[\left\langle{v}\middle|{H v}\right\rangle = \left\langle{v}\middle|{\lambda v}\right\rangle = \lambda \left\langle{v}\middle|{v}\right\rangle. \label{eq:hermitian_eigenvalue_proof_1}\] Second, using the Hermitian property of $H$, $H = H^\dagger$: \[\left\langle{v}\middle|{H v}\right\rangle = \left\langle{H^\dagger v}\middle|{v}\right\rangle = \left\langle{H v}\middle|{v}\right\rangle. \label{eq:hermitian_eigenvalue_proof_2}\] Now, substitute $H \left|{v}\right\rangle = \lambda \left|{v}\right\rangle$ into [eq:hermitian_eigenvalue_proof_2]: \[\left\langle{H v}\middle|{v}\right\rangle = \left\langle{\lambda v}\middle|{v}\right\rangle = \lambda^* \left\langle{v}\middle|{v}\right\rangle, \label{eq:hermitian_eigenvalue_proof_3}\] where $\lambda^*$ is the complex conjugate of $\lambda$.

Equating [eq:hermitian_eigenvalue_proof_1] and [eq:hermitian_eigenvalue_proof_3], we have: \[\lambda \left\langle{v}\middle|{v}\right\rangle = \lambda^* \left\langle{v}\middle|{v}\right\rangle.\] Since $\left|{v}\right\rangle$ is an eigenvector, it is non-zero, so $\left\langle{v}\middle|{v}\right\rangle = ||\left|{v}\right\rangle||^2 > 0$. Thus, we can divide both sides by $\left\langle{v}\middle|{v}\right\rangle$: \[\lambda = \lambda^*.\] This condition $\lambda = \lambda^*$ holds if and only if the imaginary part of $\lambda$ is zero, which means $\lambda$ must be a real number. Therefore, the eigenvalues of a Hermitian operator are real. ◻

Unitary Operators

Eigenvalues of Unitary Operators Have Unit Norm

Another class of special operators, unitary operators, also have a characteristic property regarding their eigenvalues: they all lie on the unit circle in the complex plane.

Theorem 3. The eigenvalues of a unitary operator have a norm of one, i.e., if $\lambda$ is an eigenvalue of a unitary operator, then $|\lambda| = 1$.

Description: This theorem explains that for a unitary operator, all its eigenvalues, when represented in the complex plane, will lie on the unit circle, meaning their absolute value is always equal to 1.

Proof. Proof. Let $U$ be a unitary operator, and let $\lambda$ be an eigenvalue of $U$ with corresponding eigenvector $\left|{v}\right\rangle \neq 0$, so $U \left|{v}\right\rangle = \lambda \left|{v}\right\rangle$. Consider the norm squared of the vector $U\left|{v}\right\rangle$, denoted as $||U\left|{v}\right\rangle||^2 = \left\langle{Uv}\middle|{Uv}\right\rangle$.

On one hand, using the eigenvalue equation, we have: \[\left\langle{Uv}\middle|{Uv}\right\rangle = \left\langle{\lambda v}\middle|{\lambda v}\right\rangle = \lambda^* \lambda \left\langle{v}\middle|{v}\right\rangle = |\lambda|^2 \left\langle{v}\middle|{v}\right\rangle = |\lambda|^2 ||\left|{v}\right\rangle||^2. \label{eq:unitary_eigenvalue_proof_1}\] On the other hand, since $U$ is unitary, by definition $U^\dagger U = I$, where $I$ is the identity operator. Therefore: \[\left\langle{Uv}\middle|{Uv}\right\rangle = \left\langle{v}\middle|{U^\dagger U v}\right\rangle = \left\langle{v}\middle|{I v}\right\rangle = \left\langle{v}\middle|{v}\right\rangle = ||\left|{v}\right\rangle||^2. \label{eq:unitary_eigenvalue_proof_2}\] Equating [eq:unitary_eigenvalue_proof_1] and [eq:unitary_eigenvalue_proof_2], we get: \[|\lambda|^2 ||\left|{v}\right\rangle||^2 = ||\left|{v}\right\rangle||^2.\] Since $\left|{v}\right\rangle \neq 0$, Since $\left|{v}\right\rangle \neq 0$, $|| \left|{v}\right\rangle ||^2 > 0$, so we can divide both sides by $|| \left|{v}\right\rangle ||^2$: \[|\lambda|^2 = 1.\] Taking the square root of both sides, we find $|\lambda| = 1$. Thus, the eigenvalues of unitary operators are complex numbers with a magnitude of 1, which can be written in the form $\lambda = e^{i\phi}$ for some real angle $\phi$. ◻

Trace of an Operator

Definition and Fundamental Properties of the Trace

The trace of an operator is a scalar value that provides important information about the operator.

Definition 8. The trace of an operator $A$, denoted as $\text{Tr}(A)$, is defined as the sum of the diagonal elements of its matrix representation in any orthonormal basis. If $A$ is represented by a matrix $[A_{ij}]$ in an orthonormal basis $\{\left|{e_i}\right\rangle\}$, then the trace is: \[\text{Tr}(A) = \sum_{i} A_{ii} = \sum_{i} \left\langle{e_i}\middle|{A e_i}\right\rangle. \label{eq:trace_definition}\]

An important property of the trace is its basis-independence. The value of the trace is the same regardless of the orthonormal basis chosen to represent the operator. Furthermore, the trace is a linear operation. For any scalars $c_1, c_2 \in \mathbb{C}$ and operators $A, B$, the trace satisfies: \[\text{Tr}(c_1 A + c_2 B) = c_1 \text{Tr}(A) + c_2 \text{Tr}(B). \label{eq:trace_linearity}\]

Cyclic Property of the Trace

The trace has a useful cyclic property, which is particularly important in manipulating expressions involving traces of operator products.

Proposition 4. For operators $A, B, C$, the trace is cyclic, meaning: \[\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB). \label{eq:trace_cyclic_property}\] More generally, for a product of $n$ operators $A_1 A_2 \cdots A_n$, the trace is invariant under cyclic permutations of the operators’ order.

Description: This proposition describes a key property of the trace operation: the trace of a product of operators remains unchanged if the order of operators is cyclically permuted. For example, Tr(ABC) = Tr(BCA) = Tr(CAB).

Trace in Terms of Eigenvalues

The trace of a normal operator has a direct relationship with its eigenvalues, providing a way to calculate the sum of eigenvalues without explicitly solving for each eigenvalue.

Theorem 5. The trace of a normal operator is equal to the sum of its eigenvalues.

Description: This theorem establishes that for a normal operator, the trace is exactly equal to the sum of all its eigenvalues. This provides a useful way to calculate the sum of eigenvalues without needing to find each eigenvalue individually.

Proof. Proof. Let $A$ be a normal operator with spectral representation $A = \sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right|$, where $\{\left|{v_i}\right\rangle\}$ is a complete orthonormal basis of eigenvectors and $\{\lambda_i\}$ are the corresponding eigenvalues. Using the definition of the trace and the spectral representation, we have: \[\begin{aligned} \text{Tr}(A) &= \text{Tr}\left(\sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right|\right) \\ &= \sum_{i} \lambda_i \text{Tr}(\left|{v_i}\right\rangle \left\langle{v_i}\right|) & \text{by linearity of trace} \\ &= \sum_{i} \lambda_i \sum_{k} \left\langle{e_k}\middle|{\left|{v_i}\right\rangle \left\langle{v_i}\right| e_k}\right\rangle & \text{using } \cref{eq:trace_definition} \text{ for any orthonormal basis } \{\left|{e_k}\right\rangle\} \\ &= \sum_{i} \lambda_i \left\langle{v_i}\middle|{v_i}\right\rangle & \text{choosing } \{\left|{e_k}\right\rangle\} = \{\left|{v_k}\right\rangle\} \\ &= \sum_{i} \lambda_i & \text{since } \left\langle{v_i}\middle|{v_i}\right\rangle = 1 \text{ for orthonormal eigenvectors}.\end{aligned}\] Thus, the trace of a normal operator is indeed the sum of its eigenvalues. ◻

Function of an Operator

Definition using Spectral Representation

For a normal operator $A$ with spectral representation $A = \sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right|$, a function of the operator $f(A)$ is defined as: \[f(A) = \sum_{i} f(\lambda_i) \left|{v_i}\right\rangle \left\langle{v_i}\right|.\] This definition is particularly useful for functions that can be expressed as power series, such as the exponential function.

Example: Exponential of Pauli-X Operator

Consider $e^{i\theta X}$, where $X = \sigma_x$. The eigenvalues of $X$ are $\lambda_1 = 1$ and $\lambda_2 = -1$ with eigenvectors $\left|{v_1}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle + \left|{1}\right\rangle)$ and $\left|{v_2}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle - \left|{1}\right\rangle)$.

Example 2. Using the spectral representation, \[\begin{aligned} e^{i\theta X} &= e^{i\theta (1)} \left|{v_1}\right\rangle \left\langle{v_1}\right| + e^{i\theta (-1)} \left|{v_2}\right\rangle \left\langle{v_2}\right| \\ &= e^{i\theta} \left|{v_1}\right\rangle \left\langle{v_1}\right| + e^{-i\theta} \left|{v_2}\right\rangle \left\langle{v_2}\right|.\end{aligned}\] Expanding in terms of $\left|{0}\right\rangle$ and $\left|{1}\right\rangle$: \[\begin{aligned} e^{i\theta X} &= e^{i\theta} \frac{1}{2} (\left|{0}\right\rangle + \left|{1}\right\rangle)(\left\langle{0}\right| + \left\langle{1}\right|) + e^{-i\theta} \frac{1}{2} (\left|{0}\right\rangle - \left|{1}\right\rangle)(\left\langle{0}\right| - \left\langle{1}\right|) \\ &= \frac{1}{2} e^{i\theta} (\left|{0}\right\rangle\left\langle{0}\right| + \left|{0}\right\rangle\left\langle{1}\right| + \left|{1}\right\rangle\left\langle{0}\right| + \left|{1}\right\rangle\left\langle{1}\right|) + \frac{1}{2} e^{-i\theta} (\left|{0}\right\rangle\left\langle{0}\right| - \left|{0}\right\rangle\left\langle{1}\right| - \left|{1}\right\rangle\left\langle{0}\right| + \left|{1}\right\rangle\left\langle{1}\right|) \\ &= \frac{1}{2} (e^{i\theta} + e^{-i\theta}) \left|{0}\right\rangle\left\langle{0}\right| + \frac{1}{2} (e^{i\theta} - e^{-i\theta}) \left|{0}\right\rangle\left\langle{1}\right| + \frac{1}{2} (e^{i\theta} - e^{-i\theta}) \left|{1}\right\rangle\left\langle{0}\right| + \frac{1}{2} (e^{i\theta} + e^{-i\theta}) \left|{1}\right\rangle\left\langle{1}\right| \\ &= \cos\theta (\left|{0}\right\rangle\left\langle{0}\right| + \left|{1}\right\rangle\left\langle{1}\right|) + i\sin\theta (\left|{0}\right\rangle\left\langle{1}\right| + \left|{1}\right\rangle\left\langle{0}\right|) \\ &= \cos\theta I + i\sin\theta X = \begin{pmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{pmatrix}.\end{aligned}\] Expanding $e^{i\theta X} = \sum_{n=0}^\infty \frac{(i\theta X)^n}{n!} = I + i\theta X + \frac{(i\theta)^2 X^2}{2!} + \frac{(i\theta)^3 X^3}{3!} + \cdots$. Since $X^2 = I$, $X^3 = X$, $X^4 = I$, etc., we can separate even and odd powers: \[\begin{aligned} e^{i\theta X} &= \left(I - \frac{\theta^2}{2!} I + \frac{\theta^4}{4!} I - \cdots\right) + i X \left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right) \\ &= \cos\theta I + i\sin\theta X = \begin{pmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{pmatrix}.\end{aligned}\] The matrix form derived using both spectral representation and series expansion is consistent.

Unitary Property of Exponential Operator

For a Hermitian operator $H$, the operator $U = e^{iH}$ is unitary.

Theorem 6. For a Hermitian operator $H$, the operator $U = e^{iH}$ is unitary.

Description: This theorem states that if we take a Hermitian operator $H$ and compute the exponential of $iH$ (where $i$ is the imaginary unit), the resulting operator $U = e^{iH}$ will always be a unitary operator. This is a fundamental connection between Hermitian and Unitary operators, especially important in quantum mechanics where time evolution is described by unitary operators derived from Hermitian Hamiltonians.

Proof. Proof. Consider $U^\dagger = (e^{iH})^\dagger = e^{-iH^\dagger} = e^{-iH}$ since $H = H^\dagger$. Then $U U^\dagger = e^{iH} e^{-iH} = e^{iH - iH} = e^0 = I$. Similarly, $U^\dagger U = e^{-iH} e^{iH} = e^{-iH + iH} = e^0 = I$. Thus, $U U^\dagger = U^\dagger U = I$, so $U = e^{iH}$ is unitary. In our example, $X = \sigma_x$ is Hermitian, so $e^{i\theta X}$ is indeed a unitary operator. ◻

Tensor Products and Composite Systems

Introduction to Tensor Products for Composite Quantum Systems

In quantum mechanics, dealing with systems composed of multiple subsystems, known as composite quantum systems, necessitates a method to describe the combined state space. This is achieved through the concept of the tensor product. The tensor product allows us to construct the Hilbert space of a composite system from the Hilbert spaces of its individual components.

Specifically, if we have two quantum systems, where the first system is described by a vector space $U$ and the second by a vector space $V$, the composite system is described by the tensor product space $U \otimes V$. If $\left|{u}\right\rangle$ is a state vector in $U$ and $\left|{v}\right\rangle$ is a state vector in $V$, then the combined state of the composite system is represented by the tensor product of these vectors, denoted as $\left|{u}\right\rangle \otimes \left|{v}\right\rangle$. In common notation, this is often simplified to $\left|{u}\right\rangle \left|{v}\right\rangle$ or $\left|{uv}\right\rangle$.

Basis of the Tensor Product Space

To construct a basis for the tensor product space $U \otimes V$, we consider the bases of the individual spaces. Let $\{\left|{u_i}\right\rangle\}_{i=1}^n$ be a basis for space $U$ and $\{\left|{v_j}\right\rangle\}_{j=1}^m$ be a basis for space $V$. Then, the set of all possible tensor products of basis vectors, $\{\left|{u_i}\right\rangle \otimes \left|{v_j}\right\rangle\}_{i=1, j=1}^{n, m}$, forms a basis for the tensor product space $U \otimes V$.

The dimension of the tensor product space is the product of the dimensions of the individual spaces: \[\dim(U \otimes V) = \dim(U) \times \dim(V) = n \times m. \label{eq:tensor_product_dimension}\]

Example 3. Consider a system composed of two qubits. Each qubit is described by a 2-dimensional Hilbert space spanned by the basis $\{\left|{0}\right\rangle, \left|{1}\right\rangle\}$. For the composite two-qubit system, the tensor product space is 4-dimensional, spanned by the basis vectors: \[\{\left|{0}\right\rangle \otimes \left|{0}\right\rangle, \left|{0}\right\rangle \otimes \left|{1}\right\rangle, \left|{1}\right\rangle \otimes \left|{0}\right\rangle, \left|{1}\right\rangle \otimes \left|{1}\right\rangle\},\] which are commonly written in shorthand as $\{\left|{00}\right\rangle, \left|{01}\right\rangle, \left|{10}\right\rangle, \left|{11}\right\rangle\}$. Here, for instance, $\left|{00}\right\rangle \equiv \left|{0}\right\rangle \otimes \left|{0}\right\rangle$ represents the state where both qubits are in the state $\left|{0}\right\rangle$.

Fundamental Properties of Tensor Products

Bilinearity of Tensor Products

The tensor product is a bilinear operation, meaning it is linear in each of its arguments. For vectors $\left|{u_1}\right\rangle, \left|{u_2}\right\rangle \in U$, $\left|{v_1}\right\rangle, \left|{v_2}\right\rangle \in V$, and scalars $c_1, c_2, d_1, d_2 \in \mathbb{C}$, bilinearity is expressed as: \[\begin{aligned} (c_1 \left|{u_1}\right\rangle + c_2 \left|{u_2}\right\rangle) \otimes \left|{v}\right\rangle &= c_1 (\left|{u_1}\right\rangle \otimes \left|{v}\right\rangle) + c_2 (\left|{u_2}\right\rangle \otimes \left|{v}\right\rangle), \label{eq:tensor_bilinearity_1} \\ \left|{u}\right\rangle \otimes (d_1 \left|{v_1}\right\rangle + d_2 \left|{v_2}\right\rangle) &= d_1 (\left|{u}\right\rangle \otimes \left|{v_1}\right\rangle) + d_2 (\left|{u}\right\rangle \otimes \left|{v_2}\right\rangle). \label{eq:tensor_bilinearity_2}\end{aligned}\] These properties are essential for manipulating linear combinations of tensor product states.

Distributivity over Vector Addition

The tensor product is distributive over vector addition, which is a direct consequence of bilinearity. For vectors $\left|{u_1}\right\rangle, \left|{u_2}\right\rangle \in U$ and $\left|{v}\right\rangle \in V$, and $\left|{u}\right\rangle \in U$ and $\left|{v_1}\right\rangle, \left|{v_2}\right\rangle \in V$: \[\begin{aligned} (\left|{u_1}\right\rangle + \left|{u_2}\right\rangle) \otimes \left|{v}\right\rangle &= \left|{u_1}\right\rangle \otimes \left|{v}\right\rangle + \left|{u_2}\right\rangle \otimes \left|{v}\right\rangle, \label{eq:tensor_distributivity_1} \\ \left|{u}\right\rangle \otimes (\left|{v_1}\right\rangle + \left|{v_2}\right\rangle) &= \left|{u}\right\rangle \otimes \left|{v_1}\right\rangle + \left|{u}\right\rangle \otimes \left|{v_2}\right\rangle. \label{eq:tensor_distributivity_2}\end{aligned}\]

Definition of Inner Product in Tensor Product Space

To define an inner product in the tensor product space $U \otimes V$, we use the inner products defined in the individual spaces $U$ and $V$. For vectors $\left|{u_1}\right\rangle, \left|{u_2}\right\rangle \in U$ and $\left|{v_1}\right\rangle, \left|{v_2}\right\rangle \in V$, the inner product in $U \otimes V$ is defined as: \[\left\langle{u_1 \otimes v_1}\middle|{u_2 \otimes v_2}\right\rangle = \left\langle{u_1}\middle|{u_2}\right\rangle \left\langle{v_1}\middle|{v_2}\right\rangle. \label{eq:tensor_product_inner_product}\] This definition ensures that if $\{\left|{u_i}\right\rangle\}$ and $\{\left|{v_j}\right\rangle\}$ are orthonormal bases for $U$ and $V$ respectively, then the set $\{\left|{u_i}\right\rangle \otimes \left|{v_j}\right\rangle\}$ forms an orthonormal basis for $U \otimes V$.

Operators on Tensor Product Spaces

Tensor Product of Operators

Operators acting on composite systems are constructed using the tensor product as well. If $A$ is a linear operator on space $U$ and $B$ is a linear operator on space $V$, the tensor product operator $A \otimes B$ is an operator on the tensor product space $U \otimes V$. Its action on simple tensor product states is defined as: \[(A \otimes B) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = (A \left|{u}\right\rangle) \otimes (B \left|{v}\right\rangle). \label{eq:tensor_product_operator_definition}\] By linearity, this definition extends to all vectors in $U \otimes V$.

Operators Acting Locally on Subsystems

In quantum mechanics, it is common to consider operators that act non-trivially on only one subsystem while acting as the identity on the other. For an operator $A$ acting on space $U$, the operator that acts as $A$ on subsystem $U$ and as the identity on subsystem $V$ is given by $A \otimes I_V$, where $I_V$ is the identity operator on $V$. Its action is: \[(A \otimes I_V) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = (A \left|{u}\right\rangle) \otimes (I_V \left|{v}\right\rangle) = (A \left|{u}\right\rangle) \otimes \left|{v}\right\rangle. \label{eq:local_operator_A_otimes_I}\] Similarly, an operator $B$ acting on space $V$ can be extended to act on $U \otimes V$ as $I_U \otimes B$, where $I_U$ is the identity operator on $U$: \[(I_U \otimes B) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = (I_U \left|{u}\right\rangle) \otimes (B \left|{v}\right\rangle) = \left|{u}\right\rangle \otimes (B \left|{v}\right\rangle). \label{eq:local_operator_I_otimes_B}\]

Commutation of Operators Acting on Different Subsystems

Operators that act on different subsystems always commute. Consider an operator $A$ acting on space $U$ and an operator $B$ acting on space $V$. Then the operators $A \otimes I_V$ and $I_U \otimes B$ acting on $U \otimes V$ commute: \[[A \otimes I_V, I_U \otimes B] = (A \otimes I_V)(I_U \otimes B) - (I_U \otimes B)(A \otimes I_V) = 0. \label{eq:commuting_operators_subsystems}\] This commutativity arises because the operators act on independent degrees of freedom. Applying $A$ on the first subsystem and then $B$ on the second yields the same result as applying $B$ first and then $A$. This can be verified by examining their action on a tensor product state $\left|{u}\right\rangle \otimes \left|{v}\right\rangle$: \[\begin{aligned} (A \otimes I_V)(I_U \otimes B) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) &= (A \otimes I_V) (\left|{u}\right\rangle \otimes B\left|{v}\right\rangle) = A\left|{u}\right\rangle \otimes B\left|{v}\right\rangle, \\ (I_U \otimes B)(A \otimes I_V) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) &= (I_U \otimes B) (A\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = A\left|{u}\right\rangle \otimes B\left|{v}\right\rangle.\end{aligned}\] Since both compositions yield the same result, their commutator is zero.

Matrix Representation of Tensor Products

When operators and vectors are represented as matrices and column vectors respectively, the tensor product operation has a corresponding matrix representation known as the Kronecker product.

If $A$ is an $n \times n$ matrix representing an operator on $U$ and $B$ is an $m \times m$ matrix representing an operator on $V$, then the operator $A \otimes B$ on $U \otimes V$ is represented by an $(nm) \times (nm)$ matrix, which is the Kronecker product of $A$ and $B$.

Given $A = [A_{ij}]$ and $B = [B_{kl}]$, their Kronecker product $A \otimes B$ is a block matrix of the form: \[A \otimes B = \begin{pmatrix} A_{11} B & A_{12} B & \cdots & A_{1n} B \\ A_{21} B & A_{22} B & \cdots & A_{2n} B \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1} B & A_{n2} B & \cdots & A_{nn} B \end{pmatrix}. \label{eq:kronecker_product_matrix}\] In this block matrix, each element $A_{ij} B$ is a block that is obtained by multiplying the entire matrix $B$ by the scalar entry $A_{ij}$ of matrix $A$.

For vectors, if $\mathbf{u} = (u_1, \ldots, u_n)^T$ is a column vector in $U$ and $\mathbf{v} = (v_1, \ldots, v_m)^T$ is a column vector in $V$, their tensor product $\mathbf{u} \otimes \mathbf{v}$ is a column vector of size $nm$ given by: \[\mathbf{u} \otimes \mathbf{v} = \begin{pmatrix} u_1 \mathbf{v} \\ u_2 \mathbf{v} \\ \vdots \\ u_n \mathbf{v} \end{pmatrix} = \begin{pmatrix} u_1 v_1 \\ u_1 v_2 \\ \vdots \\ u_1 v_m \\ u_2 v_1 \\ u_2 v_2 \\ \vdots \\ u_2 v_m \\ \vdots \\ u_n v_1 \\ u_n v_2 \\ \vdots \\ u_n v_m \end{pmatrix}. \label{eq:kronecker_product_vector}\] The components of the tensor product vector are ordered lexicographically, with the index corresponding to the first vector $\mathbf{u}$ changing more slowly than the index of the second vector $\mathbf{v}$.

Conclusion

This lecture has provided a detailed exposition of operators in quantum mechanics, with a particular emphasis on unitary, normal, and Hermitian operators, and their critical roles in the quantum mechanical framework. We have thoroughly examined the eigenvalue problem, detailing its mathematical formulation and its importance in determining eigenvectors and eigenvalues, which are essential for characterizing quantum operators and states. The spectral representation was introduced as a valuable method for expressing normal operators through their eigenvalues and eigenvectors, offering a compact and insightful perspective. We have highlighted the unique eigenvalue properties of Hermitian and unitary operators and their physical significance. Finally, we introduced tensor products as the mathematical tool for describing composite quantum systems, setting the stage for understanding multi-qubit systems and the phenomenon of quantum entanglement.

Key Takeaways:

Unitary Operators: Describe time evolution in quantum systems and represent transformations between orthonormal bases, preserving probability.
Hermitian Operators: Represent physical observables in quantum mechanics and are characterized by real eigenvalues, corresponding to measurable quantities.
Normal Operators: A broad class of operators, including Hermitian and Unitary operators, that are diagonalizable and possess orthogonal eigenvectors, simplifying their analysis and application.
Eigenvalue Problem: A fundamental mathematical problem in quantum mechanics that provides eigenvalues and eigenvectors, crucial for understanding the behavior and properties of quantum operators and states.
Spectral Representation: A powerful decomposition of normal operators in terms of their eigenvalues and eigenvectors, facilitating the definition of functions of operators and simplifying calculations.
Tensor Products: The essential mathematical construct for describing composite quantum systems, enabling the representation of multi-particle states and interactions.

Further Topics: For future lectures, we propose to delve into the following topics to build upon the foundational concepts covered today:

Applications of Spectral Representation: Exploring the use of spectral decomposition in quantum measurements and quantum state analysis.
Advanced Tensor Product Operations: Examining detailed examples of tensor product operations in quantum computing algorithms and quantum information processing.
Quantum Entanglement: Introducing the concept of quantum entanglement, its properties, and its implications for quantum technologies.
Quantum Gates and Unitary Operators: A deeper exploration of complex quantum gates, their representation as unitary operators, and their composition in quantum circuits.

--- title: "Lecture Notes on Operators and Eigenvalue Problems in Quantum Mechanics" author: "Your Name" date: "2025-02-05" format: html: toc: true # Table of Contents toc-depth: 2 code-tools: true theme: cosmo # Or "journal" for Distill-like minimalism --- # Introduction This lecture addresses essential concepts in quantum mechanics, focusing on operators and eigenvalue problems. We begin by rectifying a mistake from the previous lecture, followed by an introduction to the change of basis and unitary operators. We emphasize the importance of unitary operators in quantum mechanics, particularly in describing time evolution. Subsequently, we define and discuss various types of operators, including unitary, normal, and Hermitian operators, highlighting their properties and significance. A significant portion of this lecture is dedicated to the eigenvalue problem, covering its formulation, the secular equation, and the concept of eigenvectors. We explore the properties of eigenvectors, especially for normal operators, and introduce the spectral representation of operators. Finally, we discuss special operators such as Hermitian and unitary operators, their unique eigenvalue characteristics, the trace of an operator, functions of operators, and tensor products for composite systems. ## Correction from Previous Lecture We begin by addressing an error from the previous lecture concerning index manipulation in a demonstration. The correction is presented in these slides for clarity and ease of understanding. ## Change of Basis and Unitary Operators In vector spaces, a change of basis can be described by operators, denoted here as $R$, that satisfy the following conditions: $$R R^\dagger = I \quad \text{and} \quad R^\dagger R = I, \label{eq:unitary_condition}$$ where $I$ is the identity operator. This is analogous to rotations in linear algebra, where the inverse of a rotation $R$ is $R^{-1}$, and in this context, the adjoint operator $R^\dagger$ serves as the inverse, i.e., $R^\dagger = R^{-1}$. These operators, which include rotations and reflections, are characterized by their property of preserving the length or norm of vectors. :::: tcolorbox ::: definition **Definition 1**. A transformation $U$ is termed a **unitary transformation** or **unitary operator** if it satisfies the condition: $$U U^\dagger = U^\dagger U = I. \label{eq:unitary_operator_definition}$$ In quantum computing, such operators are often referred to simply as **unitaries**. ::: :::: Unitary operators are crucial because they preserve the norm of vectors, a property that is essential in quantum mechanics for maintaining the conservation of probabilities under transformations. ## Importance of Unitary Operators in Quantum Mechanics Unitary operators are of fundamental importance in quantum mechanics as they describe the **time evolution of quantum systems**. In the context of quantum circuits, the evolution of a quantum state as it passes through the circuit is represented by a sequence of unitary operators. Each quantum gate in a circuit corresponds to a unitary transformation. Consequently, any quantum circuit can be mathematically represented by a composition of unitary matrices. Furthermore, unitary operators can be interpreted geometrically as rotations in Hilbert space, effectively changing from one orthonormal basis to another. This perspective is valuable for visualizing and understanding the transformations applied to quantum states. # Operators in Quantum Mechanics ## Unitary Operators ### Definition and Properties :::: tcolorbox ::: definition **Definition 2**. An operator $U$ is defined as **unitary** if its adjoint $U^\dagger$ is also its inverse, satisfying the condition: $$U U^\dagger = U^\dagger U = I, \label{eq:unitary_formal_definition}$$ where $I$ represents the identity operator. ::: :::: Unitary operators are crucial in quantum mechanics as they guarantee that transformations are both **reversible** and **probability-preserving**. This reversibility ensures that quantum evolution can be traced back, and the probability preservation is essential for maintaining the probabilistic interpretation of quantum states. ### Unitary Operators as Time Evolution Operators Unitary operators play a central role in quantum mechanics by describing the **time evolution of isolated quantum systems**. The transition of a quantum state from time $t_1$ to time $t_2$ is mathematically described by the application of a unitary operator. This concept is fundamental to both quantum dynamics and quantum computation, where sequences of unitary operators represent quantum algorithms. ## Normal Operators ### Definition :::: tcolorbox ::: definition **Definition 3**. An operator $A$ is termed a **normal operator** if it commutes with its adjoint $A^\dagger$. The condition for normality is given by: $$A A^\dagger = A^\dagger A. \label{eq:normal_operator_definition}$$ ::: :::: Normal operators are significant because of their desirable spectral properties. Specifically, they are **diagonalizable**, and their eigenvectors are guaranteed to form an **orthonormal basis** for the Hilbert space on which they act. This property simplifies many theoretical and computational tasks in quantum mechanics. ## Hermitian Operators ### Definition :::: tcolorbox ::: definition **Definition 4**. An operator $H$ is defined as a **Hermitian operator** if it is equal to its adjoint, i.e., $$H = H^\dagger. \label{eq:hermitian_operator_definition}$$ ::: :::: Hermitian operators constitute an important subclass of normal operators. If an operator $H$ is Hermitian ($H = H^\dagger$), it directly follows that $H H^\dagger = H H = H^\dagger H$, thus satisfying the commutative property required for normal operators as defined in [\[eq:normal_operator_definition\]](#eq:normal_operator_definition){reference-type="ref+label" reference="eq:normal_operator_definition"}. ### Physical Observables and Hermitian Operators In quantum mechanics, a fundamental postulate is that every **physical observable** quantity---such as energy, momentum, position, and spin---is represented by a Hermitian operator. This correspondence is a cornerstone of quantum theory, linking mathematical operators to measurable physical properties. :::: tcolorbox ::: remark **Remark 1**. It is important to note that while every physical observable is represented by a Hermitian operator, the converse is not necessarily true. Not every Hermitian operator corresponds to a directly observable physical quantity. However, the set of all physical observables is always a subset of the set of Hermitian operators. ::: :::: # Eigenvalue Problem ## Introduction to the Eigenvalue Problem The **eigenvalue problem** is a cornerstone of linear algebra and quantum mechanics. It aims to identify specific vectors, termed **eigenvectors**, which, when acted upon by a linear operator, are merely scaled by a factor, known as the **eigenvalue**. Mathematically, for a given operator $A$ and a non-zero vector $\left|{v}\right\rangle$, the eigenvalue problem is formulated as finding scalars $\lambda \in \mathbb{C}$ and vectors $\left|{v}\right\rangle \neq 0$ that satisfy: $$A \left|{v}\right\rangle = \lambda \left|{v}\right\rangle. \label{eq:eigenvalue_equation}$$ In this equation, $\lambda$ is the **eigenvalue**, and $\left|{v}\right\rangle$ is the corresponding **eigenvector** of the operator $A$. ## Secular Equation and the Characteristic Polynomial To solve the eigenvalue problem, we rearrange [\[eq:eigenvalue_equation\]](#eq:eigenvalue_equation){reference-type="ref+label" reference="eq:eigenvalue_equation"} into the form: $$(A - \lambda I) \left|{v}\right\rangle = 0, \label{eq:secular_equation_operator}$$ where $I$ is the identity operator. For [\[eq:secular_equation_operator\]](#eq:secular_equation_operator){reference-type="ref+label" reference="eq:secular_equation_operator"} to have non-trivial solutions (i.e., $\left|{v}\right\rangle \neq 0$), the operator $(A - \lambda I)$ must be singular, meaning it must not have an inverse. This condition is satisfied if and only if the determinant of $(A - \lambda I)$ is zero. This condition leads to the **secular equation** or **characteristic equation**: $$\det(A - \lambda I) = 0. \label{eq:secular_equation_determinant}$$ Solving [\[eq:secular_equation_determinant\]](#eq:secular_equation_determinant){reference-type="ref+label" reference="eq:secular_equation_determinant"} for $\lambda$ yields the eigenvalues of the operator $A$. The expression $\det(A - \lambda I)$ is a polynomial in $\lambda$, known as the **characteristic polynomial**. ## Rank, Linear Dependence, and the Determinant The determinant of a square matrix is intrinsically linked to the concept of linear independence of its rows (or columns) and the **rank** of the matrix. :::: tcolorbox ::: definition **Definition 5**. The **rank** of a matrix is defined as the maximum number of linearly independent rows (or columns) in the matrix. For an $n \times n$ square matrix, the rank can range from 0 to $n$. A matrix is said to have **full rank** if its rank is equal to its dimension $n$. ::: :::: A crucial property is that a square matrix has a determinant of zero if and only if its rows (or columns) are linearly dependent, which is equivalent to the matrix not having full rank. In other words, if the rank of an $n \times n$ matrix is less than $n$, its determinant is zero. Conversely, a non-zero determinant implies that the matrix has full rank and its rows and columns are linearly independent. :::: tcolorbox ::: definition **Definition 6**. The determinant of an $n \times n$ matrix $M$, denoted as $\det(M)$, is formally defined using permutations as: $$\det(M) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^{n} M_{i, \sigma(i)}, \label{eq:determinant_definition}$$ where $S_n$ is the set of all permutations of $\{1, 2, \ldots, n\}$, and $\text{sgn}(\sigma)$ is the sign of the permutation $\sigma$. ::: :::: While [\[eq:determinant_definition\]](#eq:determinant_definition){reference-type="ref+label" reference="eq:determinant_definition"} provides a formal definition, for matrices of larger dimensions, calculating the determinant directly using this formula becomes computationally intensive due to the factorial growth of the number of permutations ($n!$). In practice, more efficient algorithms, such as Gaussian elimination or LU decomposition, are employed for determinant computation. However, for theoretical understanding, the definition in [\[eq:determinant_definition\]](#eq:determinant_definition){reference-type="ref+label" reference="eq:determinant_definition"} and its connection to linear dependence are fundamental. ## Existence of Eigenvalues and the Fundamental Theorem of Algebra The secular equation, $\det(A - \lambda I) = 0$, derived from the eigenvalue problem, is a polynomial equation in $\lambda$. For an $n \times n$ matrix $A$, this equation is a polynomial of degree $n$. The **Fundamental Theorem of Algebra** states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. Furthermore, a polynomial equation of degree $n$ has exactly $n$ complex roots, counted with multiplicity. Therefore, for any $n \times n$ matrix (representing an operator in an $n$-dimensional vector space), there exist precisely $n$ eigenvalues. These eigenvalues are complex numbers in general, although they may be real in specific cases (e.g., for Hermitian operators). Eigenvalues can also be degenerate, meaning that some eigenvalues may be repeated roots of the characteristic polynomial. ## Eigenvectors and Normalization Once the eigenvalues $\{\lambda_i\}$ are determined by solving the secular equation, the corresponding eigenvectors $\{\left|{v_i}\right\rangle\}$ can be found by solving the homogeneous system of linear equations given by [\[eq:secular_equation_operator\]](#eq:secular_equation_operator){reference-type="ref+label" reference="eq:secular_equation_operator"} for each $\lambda_i$: $$(A - \lambda_i I) \left|{v_i}\right\rangle = 0. \label{eq:eigenvector_equation}$$ For each eigenvalue $\lambda_i$, [\[eq:eigenvector_equation\]](#eq:eigenvector_equation){reference-type="ref+label" reference="eq:eigenvector_equation"} guarantees at least one non-zero solution $\left|{v_i}\right\rangle$, which is the eigenvector associated with $\lambda_i$. It is important to note that eigenvectors are defined up to a non-zero multiplicative constant. To standardize eigenvectors and for convenience in many applications, it is common practice to **normalize** them such that they have unit length, i.e., $\left\langle{v_i}\middle|{v_i}\right\rangle = 1$. ## Example: Eigenvalues and Eigenvectors of the Pauli-X Matrix To illustrate the eigenvalue problem, consider the Pauli-X matrix, a fundamental operator in quantum computing: $$\sigma_x = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \label{eq:pauli_x_matrix}$$ To find the eigenvalues of $X$, we solve the secular equation $\det(X - \lambda I) = 0$: $$\det \begin{pmatrix} -\lambda & 1 \\ 1 & -\lambda \end{pmatrix} = (-\lambda)(-\lambda) - (1)(1) = \lambda^2 - 1 = 0. \label{eq:secular_equation_pauli_x}$$ The solutions to [\[eq:secular_equation_pauli_x\]](#eq:secular_equation_pauli_x){reference-type="ref+label" reference="eq:secular_equation_pauli_x"} are $\lambda_1 = 1$ and $\lambda_2 = -1$, which are the eigenvalues of the Pauli-X matrix. To find the eigenvector $\left|{v_1}\right\rangle$ corresponding to $\lambda_1 = 1$, we solve $(X - I) \left|{v_1}\right\rangle = 0$: $$\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \label{eq:eigenvector_equation_lambda1}$$ This matrix equation yields the condition $-\alpha + \beta = 0$, or $\beta = \alpha$. Choosing $\alpha = 1$ and normalizing the resulting vector, we obtain the normalized eigenvector: $$\left|{v_1}\right\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} (\left|{0}\right\rangle + \left|{1}\right\rangle). \label{eq:eigenvector_v1_pauli_x}$$ Similarly, for the eigenvalue $\lambda_2 = -1$, we solve $(X + I) \left|{v_2}\right\rangle = 0$: $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \label{eq:eigenvector_equation_lambda2}$$ From [\[eq:eigenvector_equation_lambda2\]](#eq:eigenvector_equation_lambda2){reference-type="ref+label" reference="eq:eigenvector_equation_lambda2"}, we get $\alpha + \beta = 0$, or $\beta = -\alpha$. Choosing $\alpha = 1$ and normalizing, we find the normalized eigenvector: $$\left|{v_2}\right\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} (\left|{0}\right\rangle - \left|{1}\right\rangle). \label{eq:eigenvector_v2_pauli_x}$$ Thus, $\left|{v_1}\right\rangle$ and $\left|{v_2}\right\rangle$ are the eigenvectors of the Pauli-X matrix corresponding to the eigenvalues $1$ and $-1$, respectively. # Properties of Eigenvectors and Spectral Representation ## Eigenvectors of Normal Operators ### Orthogonality of Eigenvectors for Normal Operators A crucial property of normal operators is the orthogonality of their eigenvectors, especially when the eigenvalues are distinct. :::: tcolorbox ::: {.theorem:orthogonality_eigenvectors_normal .theorem} **Theorem 1**. *Eigenvectors of a normal operator corresponding to distinct eigenvalues are mutually orthogonal. * ::: **Description:** This theorem states that for a normal operator, if we have eigenvectors associated with different eigenvalues, those eigenvectors will always be orthogonal to each other. :::: ::: proof *Proof.* Let $A$ be a normal operator, and consider two eigenvectors $\left|{v_i}\right\rangle$ and $\left|{v_j}\right\rangle$ corresponding to distinct eigenvalues $\lambda_i$ and $\lambda_j$ respectively, such that $\lambda_i \neq \lambda_j$. By definition, we have: $$\begin{aligned} A \left|{v_i}\right\rangle &= \lambda_i \left|{v_i}\right\rangle, \label{eq:eigenvector_eq_i} \\ A \left|{v_j}\right\rangle &= \lambda_j \left|{v_j}\right\rangle. \label{eq:eigenvector_eq_j}\end{aligned}$$ Consider the inner product $\left\langle{v_i}\middle|{A v_j}\right\rangle$. Using [\[eq:eigenvector_eq_j\]](#eq:eigenvector_eq_j){reference-type="ref+label" reference="eq:eigenvector_eq_j"}, we can write: $$\left\langle{v_i}\middle|{A v_j}\right\rangle = \left\langle{v_i}\middle|{\lambda_j v_j}\right\rangle = \lambda_j \left\langle{v_i}\middle|{v_j}\right\rangle. \label{eq:inner_product_Avj}$$ Alternatively, we can use the property of the adjoint operator and the normality of $A$ ($A A^\dagger = A^\dagger A$). We know that if $\left|{v_i}\right\rangle$ is an eigenvector of a normal operator $A$ with eigenvalue $\lambda_i$, then $\left|{v_i}\right\rangle$ is also an eigenvector of $A^\dagger$ with eigenvalue $\lambda_i^*$. Thus, $A^\dagger \left|{v_i}\right\rangle = \lambda_i^* \left|{v_i}\right\rangle$. Now, consider $\left\langle{v_i}\middle|{A v_j}\right\rangle$ again, and use the adjoint property: $$\left\langle{v_i}\middle|{A v_j}\right\rangle = \left\langle{A^\dagger v_i}\middle|{v_j}\right\rangle = \left\langle{\lambda_i^* v_i}\middle|{v_j}\right\rangle = \lambda_i \left\langle{v_i}\middle|{v_j}\right\rangle. \label{eq:inner_product_Adag_vi}$$ Equating [\[eq:inner_product_Avj\]](#eq:inner_product_Avj){reference-type="ref+label" reference="eq:inner_product_Avj"} and [\[eq:inner_product_Adag_vi\]](#eq:inner_product_Adag_vi){reference-type="ref+label" reference="eq:inner_product_Adag_vi"}, we get: $$\lambda_j \left\langle{v_i}\middle|{v_j}\right\rangle = \lambda_i \left\langle{v_i}\middle|{v_j}\right\rangle.$$ Rearranging this equation yields: $$(\lambda_j - \lambda_i) \left\langle{v_i}\middle|{v_j}\right\rangle = 0.$$ Since we assumed distinct eigenvalues, $\lambda_i \neq \lambda_j$, which implies $(\lambda_j - \lambda_i) \neq 0$. Therefore, for the equation to hold, we must have: $$\left\langle{v_i}\middle|{v_j}\right\rangle = 0.$$ This demonstrates that eigenvectors corresponding to distinct eigenvalues of a normal operator are orthogonal. ◻ ::: ### Orthonormal Basis from Eigenvectors For a normal operator acting on an $n$-dimensional vector space, we can always find a set of $n$ eigenvectors that form a basis for this space. According to [1](#theorem:orthogonality_eigenvectors_normal){reference-type="ref+label" reference="theorem:orthogonality_eigenvectors_normal"}, eigenvectors corresponding to distinct eigenvalues are orthogonal. In cases where eigenvalues are degenerate (i.e., an eigenvalue has multiplicity greater than one), the eigenvectors corresponding to a degenerate eigenvalue span a subspace known as the eigenspace. Within each eigenspace, we can choose a set of orthogonal vectors, for instance, by applying the Gram-Schmidt orthonormalization process if necessary, to form an orthonormal basis for that subspace. Combining the orthonormal bases from all eigenspaces, we obtain a complete orthonormal basis for the entire $n$-dimensional vector space, consisting of eigenvectors of the normal operator. ## Spectral Representation of Normal Operators The **spectral representation**, also known as the eigenvalue decomposition, provides a way to express a normal operator in terms of its eigenvalues and eigenvectors. :::: tcolorbox ::: definition **Definition 7**. For a normal operator $A$ with a complete set of orthonormal eigenvectors $\{\left|{v_i}\right\rangle\}$ and corresponding eigenvalues $\{\lambda_i\}$, the spectral representation of $A$ is given by: $$A = \sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right| = \sum_{i} \lambda_i P_i, \label{eq:spectral_representation}$$ where $P_i = \left|{v_i}\right\rangle \left\langle{v_i}\right|$ is the projection operator onto the eigenvector $\left|{v_i}\right\rangle$. ::: :::: This representation decomposes the operator $A$ into a sum of projections onto its eigenvectors, each weighted by its corresponding eigenvalue. The spectral representation is particularly useful for understanding the action of an operator and for defining functions of operators. ## Example: Spectral Representation of the Pauli-Z Matrix Consider the Pauli-Z matrix, $\sigma_z = Z$, given by: $$Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \label{eq:pauli_z_matrix}$$ The eigenvalues of $Z$ are readily found to be $\lambda_1 = 1$ and $\lambda_2 = -1$. The eigenvector corresponding to $\lambda_1 = 1$ is $\left|{v_1}\right\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \left|{0}\right\rangle$, and the eigenvector for $\lambda_2 = -1$ is $\left|{v_2}\right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \left|{1}\right\rangle$. Both eigenvectors are already orthonormal. :::: tcolorbox ::: example **Example 1**. Using [\[eq:spectral_representation\]](#eq:spectral_representation){reference-type="ref+label" reference="eq:spectral_representation"}, the spectral representation of $Z$ is: $$\begin{aligned} Z &= \lambda_1 \left|{v_1}\right\rangle \left\langle{v_1}\right| + \lambda_2 \left|{v_2}\right\rangle \left\langle{v_2}\right| \\ &= (1) \left|{0}\right\rangle \left\langle{0}\right| + (-1) \left|{1}\right\rangle \left\langle{1}\right| \\ &= \left|{0}\right\rangle \left\langle{0}\right| - \left|{1}\right\rangle \left\langle{1}\right|.\end{aligned}$$ In matrix form, this is: $$\begin{aligned} Z &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\end{aligned}$$ which is indeed the Pauli-Z matrix. This example illustrates how the spectral representation decomposes the Pauli-Z operator in terms of its eigenvalues and the projectors onto its eigenvectors, $\left|{0}\right\rangle\left\langle{0}\right|$ and $\left|{1}\right\rangle\left\langle{1}\right|$. ::: :::: # Special Operators and their Properties ## Hermitian Operators ### Real Eigenvalues of Hermitian Operators A fundamental property of Hermitian operators is that their eigenvalues are always real. This is crucial in quantum mechanics because physical observables are represented by Hermitian operators, and their eigenvalues correspond to the possible measured values, which must be real. :::: tcolorbox ::: {.theorem:hermitian_eigenvalues_real .theorem} **Theorem 2**. *The eigenvalues of a Hermitian operator are real numbers. * ::: **Description:** This theorem demonstrates that for any Hermitian operator, the eigenvalues obtained by solving the eigenvalue equation will always be real numbers. This is a crucial property in quantum mechanics as physical observables are represented by Hermitian operators, and their measured values must be real. :::: ::: proof *Proof.* Let $H$ be a Hermitian operator, and let $\lambda$ be an eigenvalue of $H$ with corresponding eigenvector $\left|{v}\right\rangle \neq 0$. By definition, this means $H \left|{v}\right\rangle = \lambda \left|{v}\right\rangle$. To prove that $\lambda$ is real, we start by considering the expectation value of $H$ in the state $\left|{v}\right\rangle$, which can be written in two ways: First, using the eigenvalue equation: $$\left\langle{v}\middle|{H v}\right\rangle = \left\langle{v}\middle|{\lambda v}\right\rangle = \lambda \left\langle{v}\middle|{v}\right\rangle. \label{eq:hermitian_eigenvalue_proof_1}$$ Second, using the Hermitian property of $H$, $H = H^\dagger$: $$\left\langle{v}\middle|{H v}\right\rangle = \left\langle{H^\dagger v}\middle|{v}\right\rangle = \left\langle{H v}\middle|{v}\right\rangle. \label{eq:hermitian_eigenvalue_proof_2}$$ Now, substitute $H \left|{v}\right\rangle = \lambda \left|{v}\right\rangle$ into [\[eq:hermitian_eigenvalue_proof_2\]](#eq:hermitian_eigenvalue_proof_2){reference-type="ref+label" reference="eq:hermitian_eigenvalue_proof_2"}: $$\left\langle{H v}\middle|{v}\right\rangle = \left\langle{\lambda v}\middle|{v}\right\rangle = \lambda^* \left\langle{v}\middle|{v}\right\rangle, \label{eq:hermitian_eigenvalue_proof_3}$$ where $\lambda^*$ is the complex conjugate of $\lambda$. Equating [\[eq:hermitian_eigenvalue_proof_1\]](#eq:hermitian_eigenvalue_proof_1){reference-type="ref+label" reference="eq:hermitian_eigenvalue_proof_1"} and [\[eq:hermitian_eigenvalue_proof_3\]](#eq:hermitian_eigenvalue_proof_3){reference-type="ref+label" reference="eq:hermitian_eigenvalue_proof_3"}, we have: $$\lambda \left\langle{v}\middle|{v}\right\rangle = \lambda^* \left\langle{v}\middle|{v}\right\rangle.$$ Since $\left|{v}\right\rangle$ is an eigenvector, it is non-zero, so $\left\langle{v}\middle|{v}\right\rangle = ||\left|{v}\right\rangle||^2 > 0$. Thus, we can divide both sides by $\left\langle{v}\middle|{v}\right\rangle$: $$\lambda = \lambda^*.$$ This condition $\lambda = \lambda^*$ holds if and only if the imaginary part of $\lambda$ is zero, which means $\lambda$ must be a real number. Therefore, the eigenvalues of a Hermitian operator are real. ◻ ::: ## Unitary Operators ### Eigenvalues of Unitary Operators Have Unit Norm Another class of special operators, unitary operators, also have a characteristic property regarding their eigenvalues: they all lie on the unit circle in the complex plane. :::: tcolorbox ::: {.theorem:unitary_eigenvalues_unit_norm .theorem} **Theorem 3**. *The eigenvalues of a unitary operator have a norm of one, i.e., if $\lambda$ is an eigenvalue of a unitary operator, then $|\lambda| = 1$. * ::: **Description:** This theorem explains that for a unitary operator, all its eigenvalues, when represented in the complex plane, will lie on the unit circle, meaning their absolute value is always equal to 1. :::: ::: proof *Proof.* Let $U$ be a unitary operator, and let $\lambda$ be an eigenvalue of $U$ with corresponding eigenvector $\left|{v}\right\rangle \neq 0$, so $U \left|{v}\right\rangle = \lambda \left|{v}\right\rangle$. Consider the norm squared of the vector $U\left|{v}\right\rangle$, denoted as $||U\left|{v}\right\rangle||^2 = \left\langle{Uv}\middle|{Uv}\right\rangle$. On one hand, using the eigenvalue equation, we have: $$\left\langle{Uv}\middle|{Uv}\right\rangle = \left\langle{\lambda v}\middle|{\lambda v}\right\rangle = \lambda^* \lambda \left\langle{v}\middle|{v}\right\rangle = |\lambda|^2 \left\langle{v}\middle|{v}\right\rangle = |\lambda|^2 ||\left|{v}\right\rangle||^2. \label{eq:unitary_eigenvalue_proof_1}$$ On the other hand, since $U$ is unitary, by definition $U^\dagger U = I$, where $I$ is the identity operator. Therefore: $$\left\langle{Uv}\middle|{Uv}\right\rangle = \left\langle{v}\middle|{U^\dagger U v}\right\rangle = \left\langle{v}\middle|{I v}\right\rangle = \left\langle{v}\middle|{v}\right\rangle = ||\left|{v}\right\rangle||^2. \label{eq:unitary_eigenvalue_proof_2}$$ Equating [\[eq:unitary_eigenvalue_proof_1\]](#eq:unitary_eigenvalue_proof_1){reference-type="ref+label" reference="eq:unitary_eigenvalue_proof_1"} and [\[eq:unitary_eigenvalue_proof_2\]](#eq:unitary_eigenvalue_proof_2){reference-type="ref+label" reference="eq:unitary_eigenvalue_proof_2"}, we get: $$|\lambda|^2 ||\left|{v}\right\rangle||^2 = ||\left|{v}\right\rangle||^2.$$ Since $\left|{v}\right\rangle \neq 0$, Since $\left|{v}\right\rangle \neq 0$, $|| \left|{v}\right\rangle ||^2 > 0$, so we can divide both sides by $|| \left|{v}\right\rangle ||^2$: $$|\lambda|^2 = 1.$$ Taking the square root of both sides, we find $|\lambda| = 1$. Thus, the eigenvalues of unitary operators are complex numbers with a magnitude of 1, which can be written in the form $\lambda = e^{i\phi}$ for some real angle $\phi$. ◻ ::: ## Trace of an Operator ### Definition and Fundamental Properties of the Trace The **trace** of an operator is a scalar value that provides important information about the operator. :::: tcolorbox ::: definition **Definition 8**. The **trace** of an operator $A$, denoted as $\text{Tr}(A)$, is defined as the sum of the diagonal elements of its matrix representation in any orthonormal basis. If $A$ is represented by a matrix $[A_{ij}]$ in an orthonormal basis $\{\left|{e_i}\right\rangle\}$, then the trace is: $$\text{Tr}(A) = \sum_{i} A_{ii} = \sum_{i} \left\langle{e_i}\middle|{A e_i}\right\rangle. \label{eq:trace_definition}$$ ::: :::: An important property of the trace is its **basis-independence**. The value of the trace is the same regardless of the orthonormal basis chosen to represent the operator. Furthermore, the trace is a **linear operation**. For any scalars $c_1, c_2 \in \mathbb{C}$ and operators $A, B$, the trace satisfies: $$\text{Tr}(c_1 A + c_2 B) = c_1 \text{Tr}(A) + c_2 \text{Tr}(B). \label{eq:trace_linearity}$$ ### Cyclic Property of the Trace The trace has a useful cyclic property, which is particularly important in manipulating expressions involving traces of operator products. :::: tcolorbox ::: proposition **Proposition 4**. *For operators $A, B, C$, the trace is cyclic, meaning: $$\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB). \label{eq:trace_cyclic_property}$$ More generally, for a product of $n$ operators $A_1 A_2 \cdots A_n$, the trace is invariant under cyclic permutations of the operators' order.* ::: **Description:** This proposition describes a key property of the trace operation: the trace of a product of operators remains unchanged if the order of operators is cyclically permuted. For example, Tr(ABC) = Tr(BCA) = Tr(CAB). :::: ### Trace in Terms of Eigenvalues The trace of a normal operator has a direct relationship with its eigenvalues, providing a way to calculate the sum of eigenvalues without explicitly solving for each eigenvalue. :::: tcolorbox ::: {.theorem:trace_sum_eigenvalues .theorem} **Theorem 5**. *The trace of a normal operator is equal to the sum of its eigenvalues. * ::: **Description:** This theorem establishes that for a normal operator, the trace is exactly equal to the sum of all its eigenvalues. This provides a useful way to calculate the sum of eigenvalues without needing to find each eigenvalue individually. :::: ::: proof *Proof.* Let $A$ be a normal operator with spectral representation $A = \sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right|$, where $\{\left|{v_i}\right\rangle\}$ is a complete orthonormal basis of eigenvectors and $\{\lambda_i\}$ are the corresponding eigenvalues. Using the definition of the trace and the spectral representation, we have: $$\begin{aligned} \text{Tr}(A) &= \text{Tr}\left(\sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right|\right) \\ &= \sum_{i} \lambda_i \text{Tr}(\left|{v_i}\right\rangle \left\langle{v_i}\right|) & \text{by linearity of trace} \\ &= \sum_{i} \lambda_i \sum_{k} \left\langle{e_k}\middle|{\left|{v_i}\right\rangle \left\langle{v_i}\right| e_k}\right\rangle & \text{using } \cref{eq:trace_definition} \text{ for any orthonormal basis } \{\left|{e_k}\right\rangle\} \\ &= \sum_{i} \lambda_i \left\langle{v_i}\middle|{v_i}\right\rangle & \text{choosing } \{\left|{e_k}\right\rangle\} = \{\left|{v_k}\right\rangle\} \\ &= \sum_{i} \lambda_i & \text{since } \left\langle{v_i}\middle|{v_i}\right\rangle = 1 \text{ for orthonormal eigenvectors}.\end{aligned}$$ Thus, the trace of a normal operator is indeed the sum of its eigenvalues. ◻ ::: ## Function of an Operator ### Definition using Spectral Representation For a normal operator $A$ with spectral representation $A = \sum_{i} \lambda_i \left|{v_i}\right\rangle \left\langle{v_i}\right|$, a function of the operator $f(A)$ is defined as: $$f(A) = \sum_{i} f(\lambda_i) \left|{v_i}\right\rangle \left\langle{v_i}\right|.$$ This definition is particularly useful for functions that can be expressed as power series, such as the exponential function. ### Example: Exponential of Pauli-X Operator Consider $e^{i\theta X}$, where $X = \sigma_x$. The eigenvalues of $X$ are $\lambda_1 = 1$ and $\lambda_2 = -1$ with eigenvectors $\left|{v_1}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle + \left|{1}\right\rangle)$ and $\left|{v_2}\right\rangle = \frac{1}{\sqrt{2}}(\left|{0}\right\rangle - \left|{1}\right\rangle)$. :::: tcolorbox ::: example **Example 2**. Using the spectral representation, $$\begin{aligned} e^{i\theta X} &= e^{i\theta (1)} \left|{v_1}\right\rangle \left\langle{v_1}\right| + e^{i\theta (-1)} \left|{v_2}\right\rangle \left\langle{v_2}\right| \\ &= e^{i\theta} \left|{v_1}\right\rangle \left\langle{v_1}\right| + e^{-i\theta} \left|{v_2}\right\rangle \left\langle{v_2}\right|.\end{aligned}$$ Expanding in terms of $\left|{0}\right\rangle$ and $\left|{1}\right\rangle$: $$\begin{aligned} e^{i\theta X} &= e^{i\theta} \frac{1}{2} (\left|{0}\right\rangle + \left|{1}\right\rangle)(\left\langle{0}\right| + \left\langle{1}\right|) + e^{-i\theta} \frac{1}{2} (\left|{0}\right\rangle - \left|{1}\right\rangle)(\left\langle{0}\right| - \left\langle{1}\right|) \\ &= \frac{1}{2} e^{i\theta} (\left|{0}\right\rangle\left\langle{0}\right| + \left|{0}\right\rangle\left\langle{1}\right| + \left|{1}\right\rangle\left\langle{0}\right| + \left|{1}\right\rangle\left\langle{1}\right|) + \frac{1}{2} e^{-i\theta} (\left|{0}\right\rangle\left\langle{0}\right| - \left|{0}\right\rangle\left\langle{1}\right| - \left|{1}\right\rangle\left\langle{0}\right| + \left|{1}\right\rangle\left\langle{1}\right|) \\ &= \frac{1}{2} (e^{i\theta} + e^{-i\theta}) \left|{0}\right\rangle\left\langle{0}\right| + \frac{1}{2} (e^{i\theta} - e^{-i\theta}) \left|{0}\right\rangle\left\langle{1}\right| + \frac{1}{2} (e^{i\theta} - e^{-i\theta}) \left|{1}\right\rangle\left\langle{0}\right| + \frac{1}{2} (e^{i\theta} + e^{-i\theta}) \left|{1}\right\rangle\left\langle{1}\right| \\ &= \cos\theta (\left|{0}\right\rangle\left\langle{0}\right| + \left|{1}\right\rangle\left\langle{1}\right|) + i\sin\theta (\left|{0}\right\rangle\left\langle{1}\right| + \left|{1}\right\rangle\left\langle{0}\right|) \\ &= \cos\theta I + i\sin\theta X = \begin{pmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{pmatrix}.\end{aligned}$$ Expanding $e^{i\theta X} = \sum_{n=0}^\infty \frac{(i\theta X)^n}{n!} = I + i\theta X + \frac{(i\theta)^2 X^2}{2!} + \frac{(i\theta)^3 X^3}{3!} + \cdots$. Since $X^2 = I$, $X^3 = X$, $X^4 = I$, etc., we can separate even and odd powers: $$\begin{aligned} e^{i\theta X} &= \left(I - \frac{\theta^2}{2!} I + \frac{\theta^4}{4!} I - \cdots\right) + i X \left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right) \\ &= \cos\theta I + i\sin\theta X = \begin{pmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{pmatrix}.\end{aligned}$$ The matrix form derived using both spectral representation and series expansion is consistent. ::: :::: ### Unitary Property of Exponential Operator For a Hermitian operator $H$, the operator $U = e^{iH}$ is unitary. :::: tcolorbox ::: theorem **Theorem 6**. *For a Hermitian operator $H$, the operator $U = e^{iH}$ is unitary.* ::: **Description:** This theorem states that if we take a Hermitian operator $H$ and compute the exponential of $iH$ (where $i$ is the imaginary unit), the resulting operator $U = e^{iH}$ will always be a unitary operator. This is a fundamental connection between Hermitian and Unitary operators, especially important in quantum mechanics where time evolution is described by unitary operators derived from Hermitian Hamiltonians. :::: ::: proof *Proof.* Consider $U^\dagger = (e^{iH})^\dagger = e^{-iH^\dagger} = e^{-iH}$ since $H = H^\dagger$. Then $U U^\dagger = e^{iH} e^{-iH} = e^{iH - iH} = e^0 = I$. Similarly, $U^\dagger U = e^{-iH} e^{iH} = e^{-iH + iH} = e^0 = I$. Thus, $U U^\dagger = U^\dagger U = I$, so $U = e^{iH}$ is unitary. In our example, $X = \sigma_x$ is Hermitian, so $e^{i\theta X}$ is indeed a unitary operator. ◻ ::: # Tensor Products and Composite Systems ## Introduction to Tensor Products for Composite Quantum Systems In quantum mechanics, dealing with systems composed of multiple subsystems, known as **composite quantum systems**, necessitates a method to describe the combined state space. This is achieved through the concept of the **tensor product**. The tensor product allows us to construct the Hilbert space of a composite system from the Hilbert spaces of its individual components. Specifically, if we have two quantum systems, where the first system is described by a vector space $U$ and the second by a vector space $V$, the composite system is described by the **tensor product space** $U \otimes V$. If $\left|{u}\right\rangle$ is a state vector in $U$ and $\left|{v}\right\rangle$ is a state vector in $V$, then the combined state of the composite system is represented by the tensor product of these vectors, denoted as $\left|{u}\right\rangle \otimes \left|{v}\right\rangle$. In common notation, this is often simplified to $\left|{u}\right\rangle \left|{v}\right\rangle$ or $\left|{uv}\right\rangle$. ## Basis of the Tensor Product Space To construct a basis for the tensor product space $U \otimes V$, we consider the bases of the individual spaces. Let $\{\left|{u_i}\right\rangle\}_{i=1}^n$ be a basis for space $U$ and $\{\left|{v_j}\right\rangle\}_{j=1}^m$ be a basis for space $V$. Then, the set of all possible tensor products of basis vectors, $\{\left|{u_i}\right\rangle \otimes \left|{v_j}\right\rangle\}_{i=1, j=1}^{n, m}$, forms a basis for the tensor product space $U \otimes V$. The dimension of the tensor product space is the product of the dimensions of the individual spaces: $$\dim(U \otimes V) = \dim(U) \times \dim(V) = n \times m. \label{eq:tensor_product_dimension}$$ :::: tcolorbox ::: example **Example 3**. Consider a system composed of two qubits. Each qubit is described by a 2-dimensional Hilbert space spanned by the basis $\{\left|{0}\right\rangle, \left|{1}\right\rangle\}$. For the composite two-qubit system, the tensor product space is 4-dimensional, spanned by the basis vectors: $$\{\left|{0}\right\rangle \otimes \left|{0}\right\rangle, \left|{0}\right\rangle \otimes \left|{1}\right\rangle, \left|{1}\right\rangle \otimes \left|{0}\right\rangle, \left|{1}\right\rangle \otimes \left|{1}\right\rangle\},$$ which are commonly written in shorthand as $\{\left|{00}\right\rangle, \left|{01}\right\rangle, \left|{10}\right\rangle, \left|{11}\right\rangle\}$. Here, for instance, $\left|{00}\right\rangle \equiv \left|{0}\right\rangle \otimes \left|{0}\right\rangle$ represents the state where both qubits are in the state $\left|{0}\right\rangle$. ::: :::: ## Fundamental Properties of Tensor Products ### Bilinearity of Tensor Products The tensor product is a **bilinear** operation, meaning it is linear in each of its arguments. For vectors $\left|{u_1}\right\rangle, \left|{u_2}\right\rangle \in U$, $\left|{v_1}\right\rangle, \left|{v_2}\right\rangle \in V$, and scalars $c_1, c_2, d_1, d_2 \in \mathbb{C}$, bilinearity is expressed as: $$\begin{aligned} (c_1 \left|{u_1}\right\rangle + c_2 \left|{u_2}\right\rangle) \otimes \left|{v}\right\rangle &= c_1 (\left|{u_1}\right\rangle \otimes \left|{v}\right\rangle) + c_2 (\left|{u_2}\right\rangle \otimes \left|{v}\right\rangle), \label{eq:tensor_bilinearity_1} \\ \left|{u}\right\rangle \otimes (d_1 \left|{v_1}\right\rangle + d_2 \left|{v_2}\right\rangle) &= d_1 (\left|{u}\right\rangle \otimes \left|{v_1}\right\rangle) + d_2 (\left|{u}\right\rangle \otimes \left|{v_2}\right\rangle). \label{eq:tensor_bilinearity_2}\end{aligned}$$ These properties are essential for manipulating linear combinations of tensor product states. ### Distributivity over Vector Addition The tensor product is **distributive** over vector addition, which is a direct consequence of bilinearity. For vectors $\left|{u_1}\right\rangle, \left|{u_2}\right\rangle \in U$ and $\left|{v}\right\rangle \in V$, and $\left|{u}\right\rangle \in U$ and $\left|{v_1}\right\rangle, \left|{v_2}\right\rangle \in V$: $$\begin{aligned} (\left|{u_1}\right\rangle + \left|{u_2}\right\rangle) \otimes \left|{v}\right\rangle &= \left|{u_1}\right\rangle \otimes \left|{v}\right\rangle + \left|{u_2}\right\rangle \otimes \left|{v}\right\rangle, \label{eq:tensor_distributivity_1} \\ \left|{u}\right\rangle \otimes (\left|{v_1}\right\rangle + \left|{v_2}\right\rangle) &= \left|{u}\right\rangle \otimes \left|{v_1}\right\rangle + \left|{u}\right\rangle \otimes \left|{v_2}\right\rangle. \label{eq:tensor_distributivity_2}\end{aligned}$$ ### Definition of Inner Product in Tensor Product Space To define an inner product in the tensor product space $U \otimes V$, we use the inner products defined in the individual spaces $U$ and $V$. For vectors $\left|{u_1}\right\rangle, \left|{u_2}\right\rangle \in U$ and $\left|{v_1}\right\rangle, \left|{v_2}\right\rangle \in V$, the inner product in $U \otimes V$ is defined as: $$\left\langle{u_1 \otimes v_1}\middle|{u_2 \otimes v_2}\right\rangle = \left\langle{u_1}\middle|{u_2}\right\rangle \left\langle{v_1}\middle|{v_2}\right\rangle. \label{eq:tensor_product_inner_product}$$ This definition ensures that if $\{\left|{u_i}\right\rangle\}$ and $\{\left|{v_j}\right\rangle\}$ are orthonormal bases for $U$ and $V$ respectively, then the set $\{\left|{u_i}\right\rangle \otimes \left|{v_j}\right\rangle\}$ forms an orthonormal basis for $U \otimes V$. ## Operators on Tensor Product Spaces ### Tensor Product of Operators Operators acting on composite systems are constructed using the tensor product as well. If $A$ is a linear operator on space $U$ and $B$ is a linear operator on space $V$, the **tensor product operator** $A \otimes B$ is an operator on the tensor product space $U \otimes V$. Its action on simple tensor product states is defined as: $$(A \otimes B) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = (A \left|{u}\right\rangle) \otimes (B \left|{v}\right\rangle). \label{eq:tensor_product_operator_definition}$$ By linearity, this definition extends to all vectors in $U \otimes V$. ### Operators Acting Locally on Subsystems In quantum mechanics, it is common to consider operators that act non-trivially on only one subsystem while acting as the identity on the other. For an operator $A$ acting on space $U$, the operator that acts as $A$ on subsystem $U$ and as the identity on subsystem $V$ is given by $A \otimes I_V$, where $I_V$ is the identity operator on $V$. Its action is: $$(A \otimes I_V) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = (A \left|{u}\right\rangle) \otimes (I_V \left|{v}\right\rangle) = (A \left|{u}\right\rangle) \otimes \left|{v}\right\rangle. \label{eq:local_operator_A_otimes_I}$$ Similarly, an operator $B$ acting on space $V$ can be extended to act on $U \otimes V$ as $I_U \otimes B$, where $I_U$ is the identity operator on $U$: $$(I_U \otimes B) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = (I_U \left|{u}\right\rangle) \otimes (B \left|{v}\right\rangle) = \left|{u}\right\rangle \otimes (B \left|{v}\right\rangle). \label{eq:local_operator_I_otimes_B}$$ ### Commutation of Operators Acting on Different Subsystems Operators that act on different subsystems always commute. Consider an operator $A$ acting on space $U$ and an operator $B$ acting on space $V$. Then the operators $A \otimes I_V$ and $I_U \otimes B$ acting on $U \otimes V$ commute: $$[A \otimes I_V, I_U \otimes B] = (A \otimes I_V)(I_U \otimes B) - (I_U \otimes B)(A \otimes I_V) = 0. \label{eq:commuting_operators_subsystems}$$ This commutativity arises because the operators act on independent degrees of freedom. Applying $A$ on the first subsystem and then $B$ on the second yields the same result as applying $B$ first and then $A$. This can be verified by examining their action on a tensor product state $\left|{u}\right\rangle \otimes \left|{v}\right\rangle$: $$\begin{aligned} (A \otimes I_V)(I_U \otimes B) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) &= (A \otimes I_V) (\left|{u}\right\rangle \otimes B\left|{v}\right\rangle) = A\left|{u}\right\rangle \otimes B\left|{v}\right\rangle, \\ (I_U \otimes B)(A \otimes I_V) (\left|{u}\right\rangle \otimes \left|{v}\right\rangle) &= (I_U \otimes B) (A\left|{u}\right\rangle \otimes \left|{v}\right\rangle) = A\left|{u}\right\rangle \otimes B\left|{v}\right\rangle.\end{aligned}$$ Since both compositions yield the same result, their commutator is zero. ## Matrix Representation of Tensor Products When operators and vectors are represented as matrices and column vectors respectively, the tensor product operation has a corresponding matrix representation known as the **Kronecker product**. If $A$ is an $n \times n$ matrix representing an operator on $U$ and $B$ is an $m \times m$ matrix representing an operator on $V$, then the operator $A \otimes B$ on $U \otimes V$ is represented by an $(nm) \times (nm)$ matrix, which is the Kronecker product of $A$ and $B$. Given $A = [A_{ij}]$ and $B = [B_{kl}]$, their Kronecker product $A \otimes B$ is a block matrix of the form: $$A \otimes B = \begin{pmatrix} A_{11} B & A_{12} B & \cdots & A_{1n} B \\ A_{21} B & A_{22} B & \cdots & A_{2n} B \\ \vdots & \vdots & \ddots & \vdots \\ A_{n1} B & A_{n2} B & \cdots & A_{nn} B \end{pmatrix}. \label{eq:kronecker_product_matrix}$$ In this block matrix, each element $A_{ij} B$ is a block that is obtained by multiplying the entire matrix $B$ by the scalar entry $A_{ij}$ of matrix $A$. For vectors, if $\mathbf{u} = (u_1, \ldots, u_n)^T$ is a column vector in $U$ and $\mathbf{v} = (v_1, \ldots, v_m)^T$ is a column vector in $V$, their tensor product $\mathbf{u} \otimes \mathbf{v}$ is a column vector of size $nm$ given by: $$\mathbf{u} \otimes \mathbf{v} = \begin{pmatrix} u_1 \mathbf{v} \\ u_2 \mathbf{v} \\ \vdots \\ u_n \mathbf{v} \end{pmatrix} = \begin{pmatrix} u_1 v_1 \\ u_1 v_2 \\ \vdots \\ u_1 v_m \\ u_2 v_1 \\ u_2 v_2 \\ \vdots \\ u_2 v_m \\ \vdots \\ u_n v_1 \\ u_n v_2 \\ \vdots \\ u_n v_m \end{pmatrix}. \label{eq:kronecker_product_vector}$$ The components of the tensor product vector are ordered lexicographically, with the index corresponding to the first vector $\mathbf{u}$ changing more slowly than the index of the second vector $\mathbf{v}$. # Conclusion This lecture has provided a detailed exposition of operators in quantum mechanics, with a particular emphasis on unitary, normal, and Hermitian operators, and their critical roles in the quantum mechanical framework. We have thoroughly examined the eigenvalue problem, detailing its mathematical formulation and its importance in determining eigenvectors and eigenvalues, which are essential for characterizing quantum operators and states. The spectral representation was introduced as a valuable method for expressing normal operators through their eigenvalues and eigenvectors, offering a compact and insightful perspective. We have highlighted the unique eigenvalue properties of Hermitian and unitary operators and their physical significance. Finally, we introduced tensor products as the mathematical tool for describing composite quantum systems, setting the stage for understanding multi-qubit systems and the phenomenon of quantum entanglement. **Key Takeaways:** - **Unitary Operators:** Describe time evolution in quantum systems and represent transformations between orthonormal bases, preserving probability. - **Hermitian Operators:** Represent physical observables in quantum mechanics and are characterized by real eigenvalues, corresponding to measurable quantities. - **Normal Operators:** A broad class of operators, including Hermitian and Unitary operators, that are diagonalizable and possess orthogonal eigenvectors, simplifying their analysis and application. - **Eigenvalue Problem:** A fundamental mathematical problem in quantum mechanics that provides eigenvalues and eigenvectors, crucial for understanding the behavior and properties of quantum operators and states. - **Spectral Representation:** A powerful decomposition of normal operators in terms of their eigenvalues and eigenvectors, facilitating the definition of functions of operators and simplifying calculations. - **Tensor Products:** The essential mathematical construct for describing composite quantum systems, enabling the representation of multi-particle states and interactions. **Further Topics:** For future lectures, we propose to delve into the following topics to build upon the foundational concepts covered today: - **Applications of Spectral Representation:** Exploring the use of spectral decomposition in quantum measurements and quantum state analysis. - **Advanced Tensor Product Operations:** Examining detailed examples of tensor product operations in quantum computing algorithms and quantum information processing. - **Quantum Entanglement:** Introducing the concept of quantum entanglement, its properties, and its implications for quantum technologies. - **Quantum Gates and Unitary Operators:** A deeper exploration of complex quantum gates, their representation as unitary operators, and their composition in quantum circuits.